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अध्याय 6: क्रमचय और संचय (Permutations & Combinations)

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CBSE • कक्षा XI • Mathematics • अध्याय 6

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

क्रमचय और संचय (Permutations & Combinations)

In Class 11 Mathematics, "Permutations and Combinations" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

🔢 Have You Ever Wondered?

How many millions of distinct alphanumeric passwords can be generated from 8 characters, or in how many ways can a cricket selector pick a world-cup e...

How many millions of distinct alphanumeric passwords can be generated from 8 characters, or in how many ways can a cricket selector pick a world-cup eleven from 15 players? Combinatorics is the mathematical science of counting without listing.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Mathematics, "Permutations and Combinations" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Factorial notation $n!$.
Fundamental principle of counting.
Algebraic equations.

इस अध्याय के लक्ष्य

Apply Fundamental Principle of Multiplication and Addition.
Define Factorial $n!$ and compute permutations: $^nP_r = \frac{n!}{(n-r)!}$.
Compute permutations with repetition: $\frac{n!}{p_1! p_2! \dots p_k!}$.
Define and apply Combinations: $^nC_r = \frac{n!}{r!(n-r)!}$.
Apply combinatorial identities: $^nC_r = ^nC_{n-r}$ and Pascal's rule $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.

अध्याय रूपरेखा एवं प्रगति

1 1. Fundamental Principles & Factori...

2 2. Permutations: Order Matters

3 3. Combinations: Selection Only

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Fundamental Principles & Factorials

Multiplication Principle: If an event can occur in $m$ different ways, followed by another event in $n$ ways, both in succession occur in $\mathbf{m \times n}$ ways.
Factorial: $\mathbf{n! = n \times (n-1) \times \dots \times 1}$ with $0! = 1$.

2. Permutations: Order Matters

A Permutation is an arrangement in a definite order of a number of objects taken some or all at a time: $$\mathbf{^nP_r = \frac{n!}{(n-r)!}} \quad (0 \le r \le n)$$ When objects are not all distinct (e.g. letters in 'MISSISSIPPI'): $\frac{n!}{p!q!r!}$.

3. Combinations: Selection Only

A Combination is a selection of items where order is completely irrelevant: $$\mathbf{^nC_r = \frac{n!}{r!(n-r)!}} \quad \text{and} \quad \mathbf{^nP_r = r! \cdot ^nC_r}$$ Key identities: $^nC_a = ^nC_b \implies a = b \text{ or } a + b = n$; and $\mathbf{^nC_r + ^nC_{r-1} = ^{n+1}C_r}$.

चित्रात्मक व्याख्या एवं मॉडल

Permutations and Combinations Master Matrix

Conceptual framework, core mechanisms, and analytical relationships

Academic Architecture

1. Fundamental Principles & Factorials • 2. Permutations: Order Matters

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Order Sensitivity: Permutations prioritize sequence; Combinations prioritize group membership.

मुख्य बिंदु 2

Factorial Base: $0! = 1$ by combinatorial definition.

मुख्य बिंदु 3

Permutation with Repeats: Dividing total factorials by individual frequency factorials.

मुख्य बिंदु 4

Complementary Combination: $^nC_r = ^nC_{n-r}$ halving computation time.

मुख्य बिंदु 5

Pascal's Identity: Recursive formula underlying Pascal's Triangle.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

In how many ways can the letters of the word 'PERMUTATIONS' be arranged if words start with P and end with S?

उत्तर एवं व्याख्या देखें

उत्तर: Total letters = 12 (T repeats twice). Fixing P at start and S at end leaves 10 letters with 2 T's. Total ways $= \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400$.
1,814,400 ways.

If $^{n}C_9 = ^{n}C_8$, find $^{n}C_{17}$.

उत्तर एवं व्याख्या देखें

उत्तर: $^{n}C_a = ^{n}C_b \implies n = a + b = 9 + 8 = 17$. Therefore, $^{17}C_{17} = 1$.
n = 17, so 17C17 = 1.

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of exactly 3 girls?

उत्तर एवं व्याख्या देखें

उत्तर: Choose 3 girls from 4 in $^4C_3$ ways, and remaining 4 boys from 9 in $^9C_4$ ways. Total ways $= ^4C_3 \times ^9C_4 = 4 \times 126 = 504$.
504 ways.

Evaluate $\frac{8!}{6! \times 2!}$.

उत्तर एवं व्याख्या देखें

उत्तर: $\frac{8 \times 7 \times 6!}{6! \times 2} = \frac{56}{2} = 28$.
28.

How many 3-digit even numbers can be formed using digits 1, 2, 3, 4, 6, 7 if no digit is repeated?

उत्तर एवं व्याख्या देखें

उत्तर: Units place must be even: 2, 4, or 6 (3 choices). Tens place: remaining 5 choices. Hundreds place: remaining 4 choices. Total numbers $= 4 \times 5 \times 3 = 60$.
60 numbers.

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